Integral value

$5^a+b^2=3^c \implies b^2=3^c-5^a$ . So, here $b$ will be even.Consider $b=2m$

Recall that $b^2 \equiv 0,1 \pmod{3}$

$5^a+b^2 \equiv (3+2)^a + b^2 \equiv 2^a+b^2 \equiv (3-1)^a+b^2 \equiv (-1)^a +b^2 \equiv (-1)^a +0,1 \pmod{3}$ .Clearly this remainder will be zero if $a$ is odd.Consider $a=2k+1$

Now the equation becomes $5^{2k+1} +4m^2 =3^c$ .Any perfect square ends with ${0,1,4,9,5,6}$ .When it is multiplied with $4$ then ends with ${0,4,6}$ .Any power raised to $5$ always ends with $5$ .Powers of $3$ ends with $3^1=3;3^2=9;3^3=7,3^4=1$ Combining only result we can get is $4+5=9$ and $5+6=1$ .So powers of $3$ will be in the form of $c=4l,4l+2$

So, putting all those in the given equation $5^{2k+1} +4m^2=3^{4l} \\ \implies 25^k.5= 9^{2l}-(2m)^2 \\ \implies 25^k.5=(9^l+2m)(9^l-2m)$

Here $gcd(9^l+2m,9^l-2m) \neq 5$ .In the L.H.S. if $k \geq 1$ then there will more than one factors of $5$ .If $9^l-2m=1$ and $9^l+2m=25^k$ .Add this two equations $2.9^l=25^k+1$ . L.H.S will ends with $2,8$ but R.H.S. ends with $6$ .So there is no solution for $k \geq 1$ .Therefore $k=0$ and $a=1$ .

If $c=4l+2$ then the case will be similar above. So, $9^l+2m=5$ and $9^l-2m=1$ .Solving this the only values are $l=1/2$ and $m=1$ .So ,required values are $c=2$ and $b=2$ .From the second part there exists one solution but this is not integral.

So , the only values are $a=1\,\,;b=2\,\,;c=2$